Question:

Step 1: Differentiate $g(x)$: $$g'(x) = 2f'\left(\frac{x^2}{2}\right) \cdot x + f'(6 - x^2) \cdot (-2x) = 2x\left[f'\left(\frac{x^2}{2}\right) - f'(6 - x^2)\right]$$ Step 2: Analyze the sign of $g'(x)$. Since $f''(x) > 0$, $f'$ is strictly increasing. So $f'(a) > f'(b) \iff a > b$. Step 3: Determine when $f'\left(\dfrac{x^2}{2}\right) > f'(6 - x^2)$: $$\frac{x^2}{2} > 6 - x^2 \implies \frac{3x^2}{2} > 6 \implies x^2 > 4 \implies |x| > 2$$ Step 4: Determine sign of $g'(x)$ in different regions: - For $|x| > 2$: $f'\left(\frac{x^2}{2}\right) - f'(6-x^2) > 0$, so $g'(x) = 2x \cdot (\text{positive})$. Sign depends on $x$: positive for $x > 2$, negative for $x < -2$. - For $0 < |x| < 2$: $f'\left(\frac{x^2}{2}\right) - f'(6-x^2) < 0$, so $g'(x) = 2x \cdot (\text{negative})$. Sign: negative for $x > 0$, positive for $x < 0$. Step 5: Conclude monotonicity: - $g'(x) > 0$ (increasing) for $x \in (-2, 0) \cup (2, \infty)$. - $g'(x) < 0$ (decreasing) for $x \in (-\infty, -2) \cup (0, 2)$.